Parametric quadratic equation

A parametric quadratic equation differs from a normal quadratic equation in that it contains an extra parameter, often referred to as p or m. Our task is then to find out what solution the quadratic equation has depending on this parameter p.

The first example

We have the following quadratic equation with the parameter p.

$$x^2+2px+9=0.$$

What different solutions does the equation have depending on the parameter p? We can distinguish three cases - the equation has no solution in the real number domain, the equation has two identical solutions and the equation has two different solutions. We distinguish these cases by using the discriminant.

The parameter behaves as a constant, i.e., for the previous quadratic equation, a = 1, b = 2p and c = 9. Now we calculate the discriminant as we are used to:

$$D=b^2-4ac=4p^2-4\cdot1\cdot9=4p^2-36.$$

So the discriminant we came up with is D = 4p² − 36. Now we have to find out when this discriminant is positive, negative and zero.

Let's start with the simplest case, when the discriminant is zero. So we solve the equation 4p² − 36 = 0. For which p does this apply? This is a purely quadratic equation and is easy to solve:

$$\begin{eqnarray} 4p^2-36&=&0\\ p^2-9&=&0\\ p^2&=&9\\ p&=&\pm3 \end{eqnarray}$$

If the parameter p is equal to plus or minus three, then the quadratic equation has one double solution.

Now we solve when the discriminant is positive. Thus we solve the quadratic inequality

$$4p^2-36>0.$$

We already know the zero points of the function f(p) = 4p² − 36, we calculated this in the previous step. From these, we put together two intervals, one "inner" and one "outer".

$$\begin{eqnarray} I_1&=&(-\infty,-3)\cup(3, \infty)\\ I_2&=&(-3, 3) \end{eqnarray}$$

In one of these intervals the function f(p) is positive, in the other negative. We find this by plugging a point from one of the intervals into the function. The easiest way is to add a zero from the interval I₂. It is true that f(0) = 0 − 36 = −36. Thus, in the interval I₂ the function is negative, in the interval I₁ it is positive.

From this we get that in the interval I₁ the equation has two different solutions and in the interval I₂ it has no real solution. You can see the graph of the function f(p) = 4p² − 36. At the plus and minus points the square root of five is zero, at the interval I₂ it is negative, and at the interval I₁ it is positive.

Calculating the roots of the equation

Now we are left to calculate the solution to the equation in those cases where the equation has a solution. We start with the case where the equation has one double root, that is, when the discriminant is zero. This is the case when the parameter is equal to plus or minus the square root of nine.

Case one, p = 3. The equation then takes the form:

$$x^2+2\cdot3x+9=0.$$

Calculate the discriminant of this equation. It should come out to be zero:

$$D=b^2-4ac=6^2-4\cdot1\cdot9=36-36=0.$$

Calculate the root using the formula:

$$x=\frac{-b\pm\sqrt{0}}{2a}=\frac{-6}{2}=-3.$$

Case Two, p = −3. The equation takes the form:

$$x^2-2\cdot3x+9=0.$$

Calculate the discriminant:

$$D=b^2-4ac=(-6)^2-4\cdot1\cdot9=36-36=0.$$

Calculate the root:

$$x=\frac{-b\pm\sqrt{0}}{2a}=\frac{6}{2}=3.$$

In the last step, we just calculate what solutions the equation has if the parameter is from the set (−∞,−3)∪(3, ∞). We have the discriminant calculated, so we calculate the roots of the equation using the formula.

$$x_{1{,}2}=\frac{-b\pm\sqrt{4p^2-36}}{2}=\frac{-b\pm\sqrt{4(p^2-9)}}{2}=\frac{-b\pm2\sqrt{p^2-9}}{2}.$$

There is no better way to modify the expression. So these are the roots of the quadratic equation if the parameter is from the interval (−∞,−3)∪(3, ∞). If the parameter is from the interval (−3, 3), then the equation has no real solution.